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Problems #7 and #14 on pg.222-223 are graded.
#7, pg.222: Find the coordinate vector [→x]B of →x relative to the given basis B={→b1,→b2,→b3} where
→x=[8−96],
and
→b1=[1−1−3],→b2=[−349],→b3=[2−24].
Solution: We must find the weights c1,c2,c3 that solve
c1→b1+c2→b2+c3→b3=→x,
i.e. we must solve
c1[1−1−3]+c2[−349]+c3[2−24]=[8−96].
Set up the augmented matrix and compute its reduced echelon form to get
[1−328−14−2−9−3946]∼[100−1010−10013].
This yields solution c1=−1,c2=−1,c3=3. Thus we have derived the relation
(−1)[1−1−3]+(−1)[−349]+3[2−24]=[8−96],
which can be observed to be true. Therefore by the defition of coordinate vectors we see
[→x]B=[c1c2c3]=[−1−13].
#14,pg.223: The set B={1−t2,t−t2,2−t+t2} is a basis for P2. Find the coordinate vector of →p(t)=1+3t−6t2 relative to B.
Solution: We write →b1=1−t2,→b2=t−t2, and →b3=2−t+t2. We must find weights c1,c2,c3 such that
c1→b1+c2→b2+c3→b3=→p,
or in other words
c1(1−t2)+c2(t−t2)+c3(2−t+t2)=1+3t−6t2.
Simplify the left hand side by combining like terms to get
(−c1−c2+c3)t2+(c2−c3)t+(c1+2c3)=1+3t−6t2.
Equating coefficients leads to the following system of equations:
{−c1−c2+c3=−6c2−c3=3c1+2c3=1.
This system can be solved by computing the reduced echelon form of the augmented matrix:
[−1−11−601−131021]∼[10030102001−1].
Therefore we see
[→p]B=[32−1].
